The setup shown was built by the UHart ASME club to model many simultaneous principles learned in Dynamics class. When the rope is at an angle of a = 35°, sphere A (mA = 2.5 kg) has a speed vo= 0.75 m/s. The coefficient of restitution between A and wedge B (mB = 7 kg) is e = 0.8 and the length of rope 1 = 1.15 m. The spring constant has a value of k = 450 N/m and = 12⁰. A B www. 0 Determine: a.) the velocities of A and B immediately after the impact, b.) the maximum deflection of the spring, assuming A does not strike B again before this point. a

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### Problem 3

The setup shown was built by the UHart ASME club to model many simultaneous principles learned in Dynamics class.

When the rope is at an angle of \( \alpha = 35^\circ \), sphere \( A \) (\( m_A = 2.5 \) kg) has a speed \( v_0 = 0.75 \) m/s. The coefficient of restitution between \( A \) and wedge \( B \) (\( m_B = 7 \) kg) is \( e = 0.8 \) and the length of rope \( l = 1.15 \) m. The spring constant has a value of \( k = 450 \) N/m and \( \theta = 12^\circ \).

#### Determine:
a.) The velocities of \( A \) and \( B \) immediately after the impact.

b.) The maximum deflection of the spring, assuming \( A \) does not strike \( B \) again before this point.

**Hints:** You will need to carefully choose your coordinate system to find a direction that cancels the impulses between the ball and the wedge. The spring does not exert a force until it is deformed.

**Explanation of Diagram:**
The diagram includes a setup showing a sphere \( A \) attached to a rope making an angle \( \alpha = 35^\circ \) with the vertical axis. The sphere has an initial velocity \( v_0 \) directed downward. There is also a wedge \( B \) which can move horizontally and is connected to a spring with a spring constant \( k \). The horizontal line at the bottom represents the ground, and the wedge is shown to be constrained on a flat surface.

#### Diagram Elements:
- Sphere \( A \) with mass \( m_A = 2.5 \) kg
- Wedge \( B \) with mass \( m_B = 7 \) kg
- Rope with length \( l = 1.15 \) m
- Spring with constant \( k = 450 \) N/m
- Angle \( \alpha = 35^\circ \)
- Angle \( \theta = 12^\circ \)
- Coefficient of restitution \( e = 0.8 \)

(Note: The coordinate system and other relevant forces may need to be considered to solve the problem accurately.)
Transcribed Image Text:### Problem 3 The setup shown was built by the UHart ASME club to model many simultaneous principles learned in Dynamics class. When the rope is at an angle of \( \alpha = 35^\circ \), sphere \( A \) (\( m_A = 2.5 \) kg) has a speed \( v_0 = 0.75 \) m/s. The coefficient of restitution between \( A \) and wedge \( B \) (\( m_B = 7 \) kg) is \( e = 0.8 \) and the length of rope \( l = 1.15 \) m. The spring constant has a value of \( k = 450 \) N/m and \( \theta = 12^\circ \). #### Determine: a.) The velocities of \( A \) and \( B \) immediately after the impact. b.) The maximum deflection of the spring, assuming \( A \) does not strike \( B \) again before this point. **Hints:** You will need to carefully choose your coordinate system to find a direction that cancels the impulses between the ball and the wedge. The spring does not exert a force until it is deformed. **Explanation of Diagram:** The diagram includes a setup showing a sphere \( A \) attached to a rope making an angle \( \alpha = 35^\circ \) with the vertical axis. The sphere has an initial velocity \( v_0 \) directed downward. There is also a wedge \( B \) which can move horizontally and is connected to a spring with a spring constant \( k \). The horizontal line at the bottom represents the ground, and the wedge is shown to be constrained on a flat surface. #### Diagram Elements: - Sphere \( A \) with mass \( m_A = 2.5 \) kg - Wedge \( B \) with mass \( m_B = 7 \) kg - Rope with length \( l = 1.15 \) m - Spring with constant \( k = 450 \) N/m - Angle \( \alpha = 35^\circ \) - Angle \( \theta = 12^\circ \) - Coefficient of restitution \( e = 0.8 \) (Note: The coordinate system and other relevant forces may need to be considered to solve the problem accurately.)
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